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3Watt LED Recessed Ceiling Downlight - 12 Volt Question

I'm new to the world of LEDs, but have recently completed a DYI using Cree XR-E as eaves lighting outside my steel building.  Comtemplating doing the same on my personal residence and was considering the 3W 3 LED ceiling downlight ac85-265 under the soffits of my home.  These can be found in great abundance on EBAY for as low at $3.99 per unit (China made  http://www.ebay.com/itm/Warm-White-3W-3LED-Recessed-Ceiling-Downlight-ac85-265V-Bulb-Lamp-wall-lamp-/110941027707?_trksid=p5197.m1992&_trkparms=aid%3D111000%26algo%3DREC.CURRENT%26ao%3D1%26asc%3D14%26meid%3D4640763913095912456%26pid%3D100015%26prg%3D1006%26rk%3D1%26sd%3D110941027707%26  ). 

My question is this:
All the units I've looked at have attached LED resistors.  However, I only want to use a 12 Volt power supply (meanwell) placing all units in series.  What do I do with the resistors attached to the fixtures (LEDs)?  Can I remove the LED resistors and attach 12 volt power supply directly to the 3 W fixture? 

Elementary question from an elementary low voltage guy.
thanks for your responses in advance. 

frollard1 year ago
Simply put:

Get a big efficient 12 volt power supply designed to run high current leds, and wire a bunch of 12 volt (current driven) leds, or

Get mains leds. They are cheap, have a constant current driver built in on board each bulb, and you use a standard light bulb housing.

Do not remove resistors from led circuits. They are there to either limit the total current (bad for power leds for efficiency) - or to help the driver deal with the nonlinear current response.

As cheap china stuff goes, I routinely recommend dealextreme...decent quality, included shipping, etc.
http://dx.com/s/led+down
Just have to get around the poor translation to find out how to actually search for an led pot light ('led down')
iceng1 year ago
I was wondering how you came up with 300 ma .....
then I remembered seeing that picture of the heatsink :-)

4.7 ohms, 1 watt resistors are definitely a tough kludge for someone who
thinks those switching regulators are resistors.
Also as supplied they are not set up for light dimming either.

I installed a set of LED lights in a vaulted ceiling for a client using http://en.wikipedia.org/wiki/Vaulted
ten setting dial of constant currents.
Now they run the array near the low light setting.

I wonder if those eBay drivers could deliver 300 ma to three series fixtures
that's 9 LEDs in series about 33 Volts.

A
LEDs are kind of particular about the power they want. Specifically, LEDs want constant current

That lighting module, the one you have linked to, includes a constant current regulator, one that wants mains power as its input, at a voltage in the range 86 to 265 VAC. I am guessing the output of that constant current regulator is about 0.3A=300mA, and it driving a series string of 3 1-watt LEDs

Mains power, in that voltage range of 86 to 265 VAC, is available many, many, places on this planet of ours.

But for some reason, this is not good enough for you.
;-P

You want to run this lighting module, that you found a good deal on, on some other kind of power, something you are calling, "12 Volt power supply (meanwell)"

I have not heard of that brand "meanwell".

Anyway, the simple, honest, and preferred, way to do this would be to find some other constant current regulator, which can provide the same amount of constant output current, which I am guessing is around 0.3A = 300 mA, only specified for the input range of your "12 Volt power supply (meanwell)", whatever that range actually is, 10 to 15 VDC, or whatever it is.

But then the lighting module is not such a good deal anymore, if you have buy a new power regulator for each one.

Or you may be thinking that your 12 Volt meanwell has an output voltage so well regulated, that you can just assume its output is 12.000 volts DC, and then just use a resistor to limit the current to a series string of 3 LEDs. If you want to do that, I think the size of the resistor you'd want is about 5 ohms. That is of course assuming the internal series resistance of your 12 volt meanwell is like 0 ohms, or otherwise much smaller than 5 ohms.

Hopefully I can persuade this page:
http://led.linear1.org/led.wiz?VS=12.0&VF=3.6&ID=300&N=3&submit=design+my+array
to draw for you a picture-diagram of it.  I told this calculator, I had 3 LEDs, and the voltage drop across each LED was 3.6V, at 300mA.

To me this resistor idea seems kind of kludgy,
http://en.wikipedia.org/wiki/Kludge
especially considering that those lighting modules come with a perfectly good power regulator, only it's the kind made to run on mains power.

Maybe if you keep searching, you will find the Hong-Kong-mongers selling LED lighting modules made to run from 12 volts DC also?
iceng1 year ago
Welcome new member !

The units link given are 3 1W LEDs probably in series
and powered by a tiny inverter Driver constant current module
that runs on 86v-265V AC.

I see no resistor.  If you remove the driver and supply 12VDC to the red
and black wires you must add a series resistor value to be determined.

Also you cannot run the fixtures in series off 12VDC !
They will each with a series resistor need to be attached to 12VDC.

a
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